6-5.
(a) " (x, t ) = A sin (kx # !t )
"#
= $! A cos (kx $ !t )
"t
ih
"#
= $ih! A cos (kx $ !t )
"t
"2#
= $k 2 A sin (kx $ !t )
2
"x
"h 2 # 2 $ "hk 2 A
#$
=
sin (kx " !t ) % ih
2
2m #x
2m
#t
(b) " (x, t ) = A cos (kx # !t ) + iA sin (kx # !t )
ih
"#
= ih! A sin (kx $ !t ) $ i 2 h! A cos (kx $ !t )
"t
= h! A cos (kx " !t ) + ih! A sin (kx " !t )
$
h2 " 2# h2k 2 A
h 2ik 2 A
=
cos
kx
$
!
t
+
sin (kx $ !t )
(
)
2m "x 2
2m
2m
=
h2k 2
" A cos (kx $ !t ) + iA sin (kx $ !t )#&
2m %
= ih
6-9.
"#
"t
if
h2k 2
= h! it does. (Equation 6-5 with V = 0)
2m
2
(a) The ground state of an infinite well is E1 = h 2 / 8mL2 = (hc ) / 8mc 2 L2
For m = m p , L = 0.1nm : E1 =
(b) For m = m p , L = 1 fm : E1 =
(1240MeV
(
8 938.3 ! 106 eV
(1240MeV
(
2
fm )
2
)(0.1nm )
= 0.021eV
2
fm )
)
2
8 938.3 ! 106 eV (1 fm )
= 205MeV
6-10. The ground state wave function is (n = 1) ! 1 (x ) = 2 / L sin (" x / L ) (Equation 632)
The probability of finding the particle in ∆x is approximately:
P (x )"x =
2
2"x
#!x $
#!x $
sin2 %
"x =
sin2 %
&
&
L
L
' L (
' L (
(a) For
x=
2 (0.002 L ) 2 " ! L #
L
!
and $x = 0.002 L, P (x )$x =
sin %
= 0.004 sin2 = 0.004
&
2
L
2
' 2L (
2 (0.002 L ) 2 " 2! L #
2L
2!
and P (x )$x =
sin %
= 0.004 sin2
= 0.0030
(b) For x =
&
3
L
3
' 3L (
(c) For x = L and P (x )"x = 0.004 sin2 ! = 0
6-11. The second excited state wave function is (n = 3) ! 3 (x ) = 2 / L sin (3" x / L )
(Equation 6-32). The probability of finding the particle in ∆x is approximately:
P (x )$x =
2
" 3! x #
sin2 %
& $x
L
' L (
(a) For
x=
2 (0.002 L ) 2 " 3! L #
L
3!
and $x = 0.002 L, P (x )$x =
sin %
= 0.004 sin2
= 0.004
&
2
L
2
' 2L (
2L
" 6! L #
and P (x )$x = 0.004 sin2 %
= 0.004 sin2 2! = 0
(b) For x =
&
3
' 3L (
" 3! L #
= 0.004 sin2 3! = 0
(c) For x = L and P (x )$x = 0.004 sin2 %
&
' L (
6-16.
En =
h2n2
h2
and
!
E
=
E
"
E
=
n 2 + 2n + 1
n
n +1
n
2
2
8mL
8mL
(
or, "En = (2n + 1)
1/ 2
" !h #
so, L = $
%
' 8mx (
h2
hc
=
2
8mL
!
1/ 2
" ! hc #
=$
2 %
' 8mc (
2
6-20.
)
(hc )
h2
E1 =
=
2
8mL
8 mc 2 L2
( )
1/ 2
" (694.3nm )(1240eV nm ) #
%
=$
6
$
%
8
0
.
511
&
10
eV
'
(
(
)
= 0.459nm
2
(a) For an electron:
(1240MeV fm )
E1 =
2
8 (0.511MeV )(10 fm )
= 3.76 ! 103 MeV
2
(b)
(1240MeV fm )
For a proton: E1 =
2
8 (938.3Mev )(10 fm )
= 2.05MeV
(c) !E21 = 3E1 (See Problem 6-16)
For the electron: !E21 = 3E1 = 1.13 " 104 MeV
For the proton: !E21 = 3E1 = 6.15MeV
6-24. Refer to MORE section “Graphical Solution of the Finite Square Well”. If there
are only two allowed energies within the well, the highest energy E2 = V0 , the
depth of the well. From Figure 6-14, ka = ! / 2 , i.e., ka =
2mE2
h
"a =! 2
where a = 1 / 2 (1.0 fm ) = 0.5 fm and m = 939.6 MeV / c 2 for the neutron.
Substituting above, squaring, and re-arranging, we have:
2
h2
"! #
E2 = V0 = $ %
2
& 2 ' 2 939.6 MeV / c 2 (0.5 fm )
(
)
2
2
(! ) (hc )
V0 =
2
8 (939.6 MeV )(0.5 fm # 10"6 nm / fm )
2
=
(! ) (197.3eV
(
8 939.6 # 106 eV
2
nm )
)(0.5 # 10
"6
nm
V0 = 2.04 ! 108 eV = 204 MeV
+"
6-28.
px =
%h $ &
+ ! (x )') i $x (*! (s )dx
*
3
3
(Equation 6-48)
#"
L
=)
0
2
3! x # h " $ 2
3! x
sin
sin
dx
%
&
L
L ' i "x ( L
L
L
=
2h "
3! x # "
3! x # " 3!
sin
cos
$
%
$
(
L i 0&
L '&
L %' $& L
#
% dx
'
Let
3! x
= y Then x = 0 " y = 0,
L
x = L " y = 3! , and
3!
L
dx = dy " dx =
dy
L
3!
2
)
Substituting above gives:
px =
=
2h L
L i 3!
2h
L i
3!
" 3! #
L &(
) sin y cos ydy $ %'
0
3!
" sin y cos ydy
0
3!
2 h " sin2 y #
2h
=
(0 $ 0 ) = 0
%
& =
L i ' 2 (0
L i
Reconiliation: px is a vector pointing half the time in the +x direction, half in the –
x direction. Ek is a scalar proportional to v 2, hence always positive.
1/ 2
6-29. For n = 3, ! 3 = (2 / L ) sin (3" x / L )
L
(a)
x = " x (2 / L )sin2 (3! x / L )dx
0
Substituting u = 3! x / L, then x = Lu / 3! and dx = (L / 3! )du. The limits
become:
x = 0 " u = 0 and x = L " u = 3!
3!
x = (2 / L )(L / 3! )(1 / 3! ) " u sin2 udu
0
3!
2
= (2 / L )(L / 3! )
2
" u 2 u sin 2u cos 2u #
$
% $
&
4
8 (0
'4
2
= (2 / L )(1 / 3! ) (3! ) / 4 = L / 2
L
(b)
x 2 = " x 2 (2 / L )sin2 (3! x / L )dx
0
Changing the variable exactly as in (a) and noting that:
3!
3!
" u3 $ u 2 1 %
u cos 2u #
u
sin
udu
=
' & ) & * sin 2u &
(
/0
4
- 6 + 4 8,
.0
2
2
"1 1 #
We obtain x 2 = % $ ! 2 & L2 = 0.320 L2
'3 8 (
6-31.
h 2 d 2!
"
+ V (x )! (x ) = E! (x )
2m dx 2
(Equation 6-18)
1 " h d #" h d #
! (x ) = $) E & V (x )%*! (x )
2m '+ i dx (, '+ i dx (,
1
pop pop! = "% E $ V (x )#&!
2m
Multiplying by ! " and integrating over the range of x,
2
pop
+"
+"
! dx = ) ! # %' E $ V (x )&(! dx
2m
#
)!
$"
$"
p2
= !$ E # V (x )"%
2m
or
p 2 = 2m !$ E # V (x )"%
For the infinite square well V(x) = 0 wherever ! (x ) does not vanish and vice
versa.
Thus, V (x ) = 0 and p
6-32.
x2 =
L2
L2
" 2
3 2!
2
n 2! 2 h 2
! 2h2
= 2mE = 2m
= 2 for n = 1
2mL2
L
(See Problem 6-29.) And x =
1/ 2
x
!x =
p
2
% x
2
# L2
L2
L2 $
=& % 2 % '
4)
( 3 2"
! 2h2
= 2 and
L
2
x
p =0
"P =
p
% p
2
# ! 2h2
$
= & 2 % 0'
( L
)
6-33. " 0 (x ) = A0 e $ m! x
+"
x =
2
0
$ A xe
2
/ 2h
# m! x 2 / h
1/ 2
1 $
#1
= L& % 2 '
(12 2" )
= 0.181L
(See Problem 6-31)
1/ 2
2
L
2
=
!h
. And " x" p = (0.181L )(! h / L ) = 0.568h
L
1/ 4
where A0 = (m! / h# )
1/ 2
dx Letting u 2 = m! x 2 / h and x = (h / m! ) u
#"
"1
2udu = (m! / h )(2 xdx ). And thus, (m! / h ) udu = xdx; limits are unchanged.
+"
2
x = A02 (h / m! ) $ ue # u du = 0 (Note that the symmetry of V(x) would also tell
#"
us that
x = 0.)
+"
x
2
=
2
0
2 # m! x 2 / h
$Axe
dx
#"
3/ 2
2
0
= A (h / m! )
+"
2 #u2
$u e
3/ 2
2
0
du = 2 A (h / m! )
#"
3/ 2
= 2 A02 (h / m! )
6-34.
+"
2 #u2
$u e
du
#"
1/ 2
" / 4 = (m! / h" )
3/ 2
(h / m! )
" / 2 = h / (2m! )
p2 1
+ m! 2 x 2 = (n + 1 / 2 )h!. For the ground state (n = 0),
2m 2
x2 =
2
h! / 2 " p 2 / 2m
2
m!
(
)
x2 =
and
h
p2
" 2 2 = h / 2m! (See
m! m !
Problem 6-33)
h
p2
h
1"
=
or
m!
mh!
2m!
p2
1
1
1"
= # p 2 = mh!
mh!
2
2
1/ 4
6-35. (a) $ 0 (x,t ) = (m! / h" ) e # m! x
(b) px op =
h $
i $x
2
/ 2 h # i!t / 2
e
2
+!
p2 =
%h $ &
"
,#! ' 0 (x, t )(* i $x )+ ' 0 (x, t )dx
2
$% 0
1/ 4
= A0 (m! x / h" ) e # m! x / 2 h e # i!t / 2
$x
2
#2$ 0
= A0 %'("m! x / h )("m! x / h ) " m! / h &( e " m! x / 2 h e " i!t / 2
2
#x
+"
(
)
p 2 = #h 2 A02 (m! / h ) $ m! x 2 / h # 1 e # m! x
2
/h
dx
#"
+"
$ +"
%
2
2
= #h 2 A02 (m! / h )& * m! x 2 / h e # m! x / h dx # * e # m! x / 2 h dx '
#"
( #"
)
(
)
1/ 2
Letting u = (m! x / h )
p
2
2
x , then
#1 / 2
2
0
= #h A (m! / h )(m! h )
+"
$ +" 2 # u 2
%
#u2
& * u e du # * e du '
#"
( #"
)
"
$"
%
2
2
1/ 2
= #h 2 A02 (m! / h ) 2 & * u 2 e # u du # * e # u du '
0
(0
)
1/ 2
= %h 2 (m" / h! )
1/ 2
(m" / h )
# !
! $
2&
%
'
& 4
2 ')
(
= h 2 (m! / h )(1 / 2 ) = mh! / 2
6-36. " 0 (x ) = C0 e # m! x
+#
(a)
% " (x )
2
0
2
/ 2h
(Equation 6-58)
+#
dx = 1 =
$#
%C
2
0
e $ m! x
2
/h
dx
$#
2
2
= C0 $ 2 I 0 = C0 $ 2 $
= C0
1 !
with " = m# / h
2 x
!h
m"
2
1/ 4
# m! $
C0 = %
&
' "h (
+$
(b) x
2
=
&
%$
=
=
(c)
+$
2
2
x " 0 dx = & x 2
%$
m!
$ 2I2 =
"h
1 m!
2 "h
V (x ) =
m! % m! x2 / h
e
dx
#h
m!
1 "
$ 2$
with # = m! / h
"h
4 #3
" h3 1 h
=
m3! 3 2 m!
1
1
1
1 h
1
m! 2 x 2 = m! 2 x 2 = m! "
= h!
2
2
2
2 m! 4
6-37. " 1 (x ) = C1 xe # m! x
+#
(a)
&
2
/ 2h
(Equation 6-58)
+#
2
" 1 (x ) dx = 1 =
$#
&C
2
1
x 2 e $ m! x
2
/h
2
dx = C1 % 2 I 2
$#
1 !
with " = m# / h
4 "3
2
= C1 $ 2 $
2
= C1 #
1 ! h3
2 m3" 3
1/ 4
# 4m3! 3 $
C1 = %
3 &
' "h (
+$
(b)
x =
,
x "1
2
%$
3
+$
(c)
x
2
=
,x
2
"1
1/ 2
3 m3! 3
2 " h3
2
/h
dx = 0
1/ 2
2
$ 4m3! 3 %
='
3 (
) "h *
(d) V (x ) =
e % m! x
& 4m3! 3 '
dx = , x (
3 )
* #h +
%$
+$
2
%$
=
1/ 2
& 4m3! 3 '
dx = , x (
3 )
* #h +
%$
+$
e % m! x
2
/h
1/ 2
$ 4m3! 3 %
& 2I4 = '
3 (
) "h *
& 2&
x 2 dx
3 "
8 #5
where # = m! / h
" h5
3 h
=
5 5
m!
2 m!
1
1
1
3 h
3
m! 2 x 2 = m! 2 x 2 = m! 2 "
= h!
2
2
2
2 m! 4
6-42. " 0 (x ) = A0 e #! x
2
/ 2h
" 1 (x ) = A1
m! # m! x2 / 2 h
xe
h
From Equation 6-58.
Note that ! 0 is an even function of x and ! 1 is an odd function of x.
+"
It follows that
$ ! ! dx = 0
0
1
#"
6-51. (a) The probability density for the ground state is
P (x ) = ! 2 (x ) = (2 / L )sin2 " x / L.
The probability of finding the particle in the range 0 < x < L/2 is:
L/2
P=
P (x )dx =
)
0
L/3
(b) P =
)
P (x )dx =
0
2L
L!
! /2
2L
L!
! /3
2
udu =
2 "!
# 1
$ 0& =
where u = ! x / L
%
!'4
( 2
2
udu =
2 " ! sin 2! / 3 # 1
3
$
= $
= 0.195
%
&
!'6
4
( 3 4!
) sin
0
) sin
0
(Note: 1/3 is the classical result.)
3L / 4
(c) P =
)
0
2L
P (x )dx =
L!
3! / 4
)
sin2 udu =
0
(Note: 3/4 is the classical result.)
2 " 3! sin 3! / 2 # 3 1
$
& = 4 + 2! = 0.909
! %' 8
4
(